Andrei R. Andrei R. - 1 year ago 225
C++ Question

struct to/from std::tuple conversion

Assuming I have

with same type layout:

struct MyStruct { int i; bool b; double d; }
using MyTuple = std::tuple<int,bool,double>;

Is there any standartized way to cast one to another?

P.S. I know that trivial memory copying can do the trick, but it is alignment and implementation dependent

Answer Source

Unfortunately there is no automatic way to do that, BUT an alternative is adapt the struct to Boost.Fusion sequence. You do this once and for all for each new class.

#include <boost/fusion/adapted/struct/adapt_struct.hpp>
struct MyStruct { int i; bool b; double d; }

    (int, i)
    (bool, b)
    (double, d)

The use MyStruct as if it where a Fusion.Sequence (it fits generically almost everywhere you already use std::tuple<...>, if you make those functions generic.) As a bonus you will not need to copy your data members at all.

If you really need to convert to std::tuple, after "Fusion-adapting" you can do this:

#include <boost/fusion/adapted/std_tuple.hpp>
#include <boost/fusion/algorithm/iteration/for_each.hpp>
#include <boost/fusion/algorithm/transformation/zip.hpp>
auto to_tuple(MyStruct const& ms){
   std::tuple<int, bool, double> ret;
   auto z = zip(ret, ms);
   boost::fusion::for_each(z, [](auto& ze){get<0>(ze) = get<1>(ze);});
   // or use boost::fusion::copy
   return ret;

The truth is that std::tuple is a half-backed feature. It is like having STD containers and no algorithms. Fortunatelly we have #include <boost/fusion/adapted/std_tuple.hpp> that allows us to do amazing things.

Full code:

By including the std_tuple.hpp header from Boost.Fusion std::tuple is automatically adapted to a Boost.Fusion sequence, thus the following is possible by using Boost.Fusion as a bridge between your struct and std::tuple:

#include <iostream>
#include <string>
#include <tuple>

#include <boost/fusion/adapted/struct/adapt_struct.hpp>
#include <boost/fusion/algorithm/auxiliary/copy.hpp>
#include <boost/fusion/adapted/std_tuple.hpp>

struct foo
  std::string a, b, c;
  int d, e, f;

    (std::string, a)
    (std::string, b)
    (std::string, c)
    (int, d)
    (int, e)
    (int, f)

template<std::size_t...Is, class Tup>
foo to_foo_aux(std::index_sequence<Is...>, Tup&& tup) {
  using std::get;
  return {get<Is>(std::forward<Tup>(tup))...};
template<class Tup>
foo to_foo(Tup&& tup) {
  using T=std::remove_reference_t<Tup>;
  return to_foo_aux(

auto to_tuple_aux( std::index_sequence<Is...>, foo const& f ) {
  using boost::fusion::at_c;
  return std::make_tuple(at_c<Is>(f)...);
auto to_tuple(foo const& f){
  using T=std::remove_reference_t<foo>;
  return to_tuple_aux(

int main(){

    foo f{ "Hello", "World", "!", 1, 2, 3 };

    std::tuple<std::string, std::string, std::string, int, int, int> dest = to_tuple(f);
    // boost::fusion::copy(f, dest); // also valid  but less general than constructor

    std::cout << std::get<0>(dest) << ' ' << std::get<1>(dest) << std::get<2>(dest) << std::endl;
    std::cout << at_c<0>(dest) << ' ' << at_c<1>(dest) << at_c<2>(dest) << std::endl; // same as above

    foo f2 = to_foo(dest);

    std::cout << at_c<0>(f2) << ' ' << at_c<1>(f2) << at_c<2>(f2) << std::endl;

I will not recommend reinterpret_cast<std::tuple<...>&>(mystructinstance.i) because that will result in negative votes and it is not portable.

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